# integer solutions to this system of equations

Consider the following system of two equations: $$x + y + z = 24 n + 3$$ $$x y z = 576 n³ + 216 n^2 + 27 n - 25$$

For what value of $n$ are there integer solutions $x$, $y$, $z$?

Can someone help me to write down a code for the above!

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I assume that $x,y,z,n$ are nonnegative.

This system is equivalent to the equation: $(x+y+z)^3 = 24xzy + 627$. By AM-GM inequality, we have $$27xyz \leq (x+y+z)^3 = 24xzy + 627,$$ implying that $xyz \leq 209$. That is, there are only a finite number of solutions, which are easy to find by brute-force.

ADDED. One can also apply AM-GM in the other direction: $$(x+y+z)^3 = 24xzy + 627 \leq \frac{24}{27}(x+y+z)^3 + 627,$$ implying that $(x+y+z)^3 \leq 5643$ and thus $24n+3=x+y+z\leq 17$. This gives $n=0$ as the only possible option, however it would mean $xyz=-25$, a contradiction.

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Very nice, under the assumption $x, y, z, n \in \mathbb{Z}^{*+}$.

What can be said for negative solutions ?

Max gives us a shortcut allowing us to use brute force to check for (positive) roots of this system. But what is the solution ?

First, let's see how Max's shortcut can be derived in Sage :

sage: var("x, y, z, p")
(x, y, z, p)


Our initial equations are :

sage: E1=x+y+z==24*p+3
sage: E2=x*y*z==576*p^3+216*p^2+27*p-25


Let's solve E2, keeping only real roots, but with no consideration for their integrity

sage: E2.solve(p)
[p == -1/16*(8/9*x*y*z + 209/9)^(1/3)*(I*sqrt(3) + 1) - 1/8, p == -1/16*(8/9*x*y*z + 209/9)^(1/3)*(-I*sqrt(3) + 1) - 1/8, p == 1/24*(24*x*y*z + 627)^(1/3) - 1/8]


Substitute in E1 :

sage: E1.subs(E2.solve(p))^3
(x + y + z)^3 == 24*x*y*z + 627


AM-GM inequality gives us :

sage: I1=(x+y+z)/3>=(x*y*z)^(1/3)

sage: ((I1*3)^3)
(x + y + z)^3 >= 27*x*y*z


sage: (((I1*3)^3).subs(E1.subs(E2.solve(p))^3))
24*x*y*z + 627 >= 27*x*y*z


which results in:

sage: I2=(((I1*3)^3).subs(E1.subs(E2.solve(p))^3)).solve(x*y*z)
[[x*y*z - 209 == 0], [-x*y*z + 209 > 0]]


This upper limit allows us to program a "brute-force" search :

sage: %%time
....: L=[]
....: for a in (1..209):
....:     for b in (1..209//a):
....:         for c in (1..209//(a*b)):
....:             D1={x:a, y:b, z:c}
....:             S1=E1.subs(D1).solve_diophantine(solution_dict=True)
....:             if len(S1)>0:
....:                 L2=[D1.update(s) for s in S1 if bool(E2.subs(D1).subs(S1))]
....:                 if len(L2)>0: L.append(L2)
....:
CPU times: user 34.3 s, sys: 16 ms, total: 34.4 s
Wall time: 34.4 s


But :

sage: L
[]


The original system doesn't seem to have solutions. A check seems in order...

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Thanks to the symmetry, you can also assume that $a\geq b\geq c$ to save some time. But in fact, the solutions can be obtained without any brute-force, if we apply AM-GM in the other direction -- see update in my answer.